Very nice, Richard. We shouldn't be so quick to get the hammer if we think
we've seen a nail!
Bobby
On Fri, 04 Nov 2011 05:59:20 -0500, Richard Fateman
<fateman at cs.berkeley.edu> wrote:
> On 11/3/2011 1:59 AM, Bob Hanlon wrote:
>> Use higher precision.
> .. so say you all...
> Well, that's the brute force method, and one reason to recommend it is
> it doesn't require much thought.
>
> On the other hand, the person asking the question has an apparently
> large expression which he knows has a double zero at n=1/2 and he wants
> to know the behavior of the expression from 0.35 to 0.53. Namely,
> around that zero.
>
> Let us call that expression p. A non-brute force, but mostly automatic
> method is to note that (since p is a polynomial of degree 29) it is
> EXACTLY equal to s, where
>
> s = Normal[Series[p, {n, 1/2, 29}]]
>
> A plot of s can be done in ordinary float arithmetic and looks to the
> eye just like the plot of p, done with high working precision. Unique to
> computer algebra systems, it is also possible to look at the structure
> of s, and note that expanded around n=1/2 it is mighty sparse.
>
> Let k = n-1/2 . The expression being plotted is then
>
> 14 k^2 - 1296 k^5 + 1451188224 k^11 - 2002639749120 k^14 +
> 598075300380672 k^17 - 83414577743659008 k^20 +
> 3977900382788517888 k^23 - 113721152119718805504 k^26 +
> 1516282028262917406720 k^29
>
>
> Sometimes a better way to format expressions for evaluation is to use
> HornerForm. In this case it is
>
>
> k^2 (14 +
> k^3 (-1296 + ....
>
> Evaluating this expression at a value for k requires computing
> k^2,k^3,k^6,and some other arithmetic for a total of about 18 arithmetic
> operations, which can be done in double-float.
> The real win here is that you can show how much more insight you might
> get from using computer algebra.
>
> RJF
>
--
DrMajorBob at yahoo.com